64*5x-125*4x>0

To solve the inequality \(64 \cdot 5^x - 125 \cdot 4^x > 0\), we first express \(4^x\) in terms of \(5^x\) for easier manipulation. Since \(4^x = (2^2)^x = (2^x)^2\), we can note that \(64 = 4^3\) and \(125 = 5^3\), allowing us to rewrite the inequality as: \[ 64 \cdot 5^x = 4^3 \cdot 5^x \] \[ 125 \cdot 4^x = 5^3 \cdot 4^x \] Thus, we rewrite our original inequality as: \[ 4^3 \cdot 5^x - 5^3 \cdot 4^x > 0 \] Now we can take out common factors for simplification. It may be more insightful to express \(4^x\) in terms of \(5^x\). Noticing that \(4^x = (2^2)^x = (2^x)^2\) and \(5 = (2.5)^x\), we can convert everything to a common base if necessary, but here let’s factor it differently: Let \(y = 5^x\) and note \(4^x = \left(\frac{4}{5}\right)^x \cdot 5^x\). Thus, we can rewrite the inequality: \[ 64y - 125 \left(\frac{4}{5}\right)^x y > 0 \] This leads us to: \[ y(64 - 125 \left(\frac{4}{5}\right)^x) > 0 \] Next, we determine the critical points by solving: \[ 64 - 125 \left(\frac{4}{5}\right)^x = 0 \] This simplifies to: \[ 125 \left(\frac{4}{5}\right)^x = 64 \] \[ \left(\frac{4}{5}\right)^x = \frac{64}{125} \] Taking logarithms of both sides, we have: \[ x \log\left(\frac{4}{5}\right) = \log\left(\frac{64}{125}\right) \] Now we can solve for \(x\): \[ x = \frac{\log\left(\frac{64}{125}\right)}{\log\left(\frac{4}{5}\right)} \] Since \(\log\left(\frac{64}{125}\right) = \log(64) - \log(125) = 3\log(4) - 3\log(5) = 3(\log(4) - \log(5))\), and knowing \( \log\left(\frac{4}{5}\right) = \log(4) - \log(5) \): Thus: \[ x = \frac{3(\log(4) - \log(5))}{\log(4) - \log(5)} = 3 \] Now we test the intervals created by the critical point \(x = 3\). 1. For \(x < 3\): choose \(x = 0\): \[ 64 \cdot 5^0 - 125 \cdot 4^0 = 64 - 125 = -61 < 0 \] 2. For \(x > 3\): choose \(x = 4\): \[ 64 \cdot 5^4 - 125 \cdot 4^4 = 64 \cdot 625 - 125 \cdot 256 = 40000 - 32000 = 8000 > 0 \] Thus, the solution to the inequality \(64 \cdot 5^x - 125 \cdot 4^x > 0\) is: \[ x > 3 \] In interval notation, the solution is: \[ (3, \infty) \]